Oxidation is loss of an electron (from an element) at the anode (this means the electrons are building up on this terminal )
eg
Cu(s) (reductant) -----> Cu 2+(aq) + 2 e- electrons have been lost from the reactant, so now they appear as the product and the element has become a cation....the copper atom is oxidised
Reduction is gain of an electron at the cathode (this means the cation will gain an electron and become a neutral atom. For most neutral atoms of metals will be solid)
Ag+(aq)(oxidant) + e- ------> Ag(s)
(electron is gained so it appears as the reactant, it balances the charge on the ion)
(an electron is gained to reduce each silver ion)
The oxidant has been reduced and the reductant has been oxidised
OXIDATION STATES (OXIDATION NUMBERS)
This page explains what oxidation states (oxidation numbers) are and how to calculate them and make use of them.
Oxidation states are straightforward to work out and to use, but it is quite difficult to define what they are in any quick way. Explaining what oxidation states (oxidation numbers) are Oxidation states simplify the whole process of working out what is being oxidised and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be helpful if you knew about:
oxidation and reduction in terms of electron transfer
electron-half-equations
Note: If you aren't sure about either of these things, you might want to look at the pages on redox definitions and electron-half-equations. It would probably be best to read on and come back to these links if you feel you need to.
We are going to look at some examples from vanadium chemistry. If you don't know anything about vanadium, it doesn't matter in the slightest.
Vanadium forms a number of different ions - for example, V2+ and V3+. If you think about how these might be produced from vanadium metal, the 2+ ion will be formed by oxidising the metal by removing two electrons:
The vanadium is now said to be in an oxidation state of +2.
Removal of another electron gives the V3+ ion:
The vanadium now has an oxidation state of +3.
Removal of another electron gives a more unusual looking ion, VO2+.
The vanadium is now in an oxidation state of +4. Notice that the oxidation state isn't simply counting the charge on the ion (that was true for the first two cases but not for this one).
The positive oxidation state is counting the total number of electrons which have had to be removed - starting from the element.
It is also possible to remove a fifth electron to give another ion (easily confused with the one before!). The oxidation state of the vanadium is now +5.
Every time you oxidise the vanadium by removing another electron from it, its oxidation state increases by 1.
Fairly obviously, if you start adding electrons again the oxidation state will fall. You could eventually get back to the element vanadium which would have an oxidation state of zero.
What if you kept on adding electrons to the element? You can't actually do that with vanadium, but you can with an element like sulphur.
The sulphur has an oxidation state of -2. Summary Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state.
Oxidation involves an increase in oxidation state Reduction involves a decrease in oxidation state
Recognising this simple pattern is the single most important thing about the concept of oxidation states. If you know how the oxidation state of an element changes during a reaction, you can instantly tell whether it is being oxidised or reduced without having to work in terms of electron-half-equations and electron transfers
. Working out oxidation states
You don't work out oxidation states by counting the numbers of electrons transferred. It would take far too long. Instead you learn some simple rules, and do some very simple sums!
The oxidation state of an uncombined element is zero. That's obviously so, because it hasn't been either oxidised or reduced yet! This applies whatever the structure of the element - whether it is, for example, Xe or Cl2 or S8, or whether it has a giant structure like carbon or silicon.
The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state. Remember that fluorine is the most electronegative element with oxygen second.
Some elements almost always have the same oxidation states in their compounds:
>
element
usual oxidation state
exceptions
|| Group 1 metals || always +1 || ||
|| Group 2 metals || always +2 || ||
|| Oxygen || usually -2 || except in peroxides and F2O (see below) ||
|| Hydrogen || usually +1 || except in metal hydrides where it is -1 (see below) ||
|| Fluorine || always -1 || ||
|| Chlorine || usually -1 || except in compounds with O or F (see below) ||
The reasons for the exceptions Hydrogen in the metal hydrides
Metal hydrides include compounds like sodium hydride, NaH. In this, the hydrogen is present as a hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion - in this case, -1.
Alternatively, you can think of it that the sum of the oxidation states in a neutral compound is zero. Since Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0). Oxygen in peroxides
Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound and so the sum of the oxidation states of the hydrogen and oxygen must be zero.
Since each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it. Oxygen in F2O
The problem here is that oxygen isn't the most electronegative element. The fluorine is more electronegative and has an oxidation state of -1. In this case, the oxygen has an oxidation state of +2. Chlorine in compounds with fluorine or oxygen
There are so many different oxidation states that chlorine can have in these, that it is safer to simply remember that the chlorine doesn't have an oxidation state of -1 in them, and work out its actual oxidation state when you need it. You will find an example of this below. Warning! Don't get too bogged down in these exceptions. In most of the cases you will come across, they don't apply!Examples of working out oxidation states What is the oxidation state of chromium in Cr2+?
That's easy! For a simple ion like this, the oxidation state is the charge on the ion - in other words: +2 (Don't forget the + sign.) What is the oxidation state of chromium in CrCl3?
This is a neutral compound so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1. If the oxidation state of chromium is n:
n + 3(-1) = 0
n = +3 (Again, don't forget the + sign!) What is the oxidation state of chromium in Cr(H2O)63+?
This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia.
The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr3+. The oxidation state is +3. What is the oxidation state of chromium in the dichromate ion, Cr2O72-?
The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present.
2n + 7(-2) = -2
n = +6
||
Warning: Because these are simple sums it is tempting to try to do them in your head. If it matters (like in an exam) write them down using as many steps as you need so that there is no chance of making careless mistakes. Your examiners aren't going to be impressed by your mental arithmetic - all they want is the right answer! If you want some more examples to practice on, you will find them in most text books, including my chemistry calculations book.
What is the oxidation state of copper in CuSO4?
Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulphur) whose oxidation states can both change.
The only way around this is to know some simple chemistry! There are two ways you might approach it. (There might be others as well, but I can't think of them at the moment!)
You might recognise this as an ionic compound containing copper ions and sulphate ions, SO42-. To make an electrically neutral compound, the copper must be present as a 2+ ion. The oxidation state is therefore +2.
You might recognise the formula as being copper(II) sulphate. The "(II)" in the name tells you that the oxidation state is 2 (see below).
You will know that it is +2 because you know that metals form positive ions, and the oxidation state will simply be the charge on the ion.
Using oxidation states In naming compounds
You will have come across names like iron(II) sulphate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe2+ and Fe3+ ions.
This can also be extended to the negative ion. Iron(II) sulphate is FeSO4. There is also a compound FeSO3 with the old name of iron(II) sulphite. The modern names reflect the oxidation states of the sulphur in the two compounds.
The sulphate ion is SO42-. The oxidation state of the sulphur is +6 (work it out!). The ion is more properly called the sulphate(VI) ion.
The sulphite ion is SO32-. The oxidation state of the sulphur is +4 (work that out as well!). This ion is more properly called the sulphate(IV) ion. The ate ending simply shows that the sulphur is in a negative ion.
So FeSO4 is properly called iron(II) sulphate(VI), and FeSO3 is iron(II) sulphate(IV). In fact, because of the easy confusion between these names, the old names sulphate and sulphite are normally still used in introductory chemistry courses.
Note: Even these aren't the full name! The oxygens in the negative ions should also be identified. FeSO4 is properly called iron(II) tetraoxosulphate(VI). It all gets a bit out of hand for everyday use for common ions.
Using oxidation states to identify what's been oxidised and what's been reduced
This is easily the most common use of oxidation states.
Remember: Oxidation involves an increase in oxidation state Reduction involves a decrease in oxidation state
In each of the following examples, we have to decide whether the reaction involves redox, and if so what has been oxidised and what reduced. Example 1:
This is the reaction between magnesium and hydrochloric acid or hydrogen chloride gas:
Have the oxidation states of anything changed? Yes they have - you have two elements which are in compounds on one side of the equation and as uncombined elements on the other. Check all the oxidation states to be sure:.
The magnesium's oxidation state has increased - it has been oxidised. The hydrogen's oxidation state has fallen - it has been reduced. The chlorine is in the same oxidation state on both sides of the equation - it hasn't been oxidised or reduced. Example 2:
The reaction between sodium hydroxide and hydrochloric acid is:
Checking all the oxidation states:
Nothing has changed. This isn't a redox reaction. Example 3:
This is a sneaky one! The reaction between chlorine and cold dilute sodium hydroxide solution is:
Obviously the chlorine has changed oxidation state because it has ended up in compounds starting from the original element. Checking all the oxidation states shows:
The chlorine is the only thing to have changed oxidation state. Has it been oxidised or reduced? Yes! Both! One atom has been reduced because its oxidation state has fallen. The other has been oxidised.
This is a good example of a disproportionation reaction. A disproportionation reaction is one in which a single substance is both oxidised and reduced. Using oxidation states to work out reacting proportions
This is sometimes useful where you have to work out reacting proportions for use in titration reactions where you don't have enough information to work out the complete ionic equation.
Remember that each time an oxidation state changes by one unit, one electron has been transferred. If one substance's oxidation state in a reaction falls by 2, that means that it has gained 2 electrons.
Something else in the reaction must be losing those electrons. Any oxidation state fall by one substance must be accompanied by an equal oxidation state increase by something else.
This example is based on information in an old AQA A' level question.
Ions containing cerium in the +4 oxidation state are oxidising agents. (They are more complicated than just Ce4+.) They can oxidise ions containing molybdenum from the +2 to the +6 oxidation state (from Mo2+ to MoO42-). In the process the cerium is reduced to the +3 oxidation state (Ce3+). What are the reacting proportions?
The oxidation state of the molybdenum is increasing by 4. That means that the oxidation state of the cerium must fall by 4 to compensate.
But the oxidation state of the cerium in each of its ions only falls from +4 to +3 - a fall of 1. So there must obviously be 4 cerium ions involved for each molybdenum ion.
The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion.
Or to take a more common example involving iron(II) ions and manganate(VII) ions . . .
A solution of potassium manganate(VII), KMnO4, acidified with dilute sulphuric acid oxidises iron(II) ions to iron(III) ions. In the process, the manganate(VII) ions are reduced to manganese(II) ions. Use oxidation states to work out the equation for the reaction.
The oxidation state of the manganese in the manganate(VII) ion is +7. The name tells you that, but work it out again just for the practice!
In going to manganese(II) ions, the oxidation state of manganese has fallen by 5. Every iron(II) ion that reacts, increases its oxidation state by 1. That means that there must be five iron(II) ions reacting for every one manganate(VII) ion.
The left-hand side of the equation will therefore be: MnO4- + 5Fe2+ + ?
The right-hand side will be: Mn2+ + 5Fe3+ + ?
After that you will have to make guesses as to how to balance the remaining atoms and the charges. In this case, for example, it is quite likely that the oxygen will end up in water. That means that you need some hydrogen from somewhere.
That isn't a problem because you have the reaction in acid solution, so the hydrogens could well come from hydrogen ions.
Eventually, you will end up with this:
Personally, I would much rather work out these equations from electron-half-equations!
Questions to test your understanding
If this is the first set of questions you have done, please read the introductory page before you start. You will need to use the BACK BUTTON on your browser to come back here afterwards. questions on oxidation states answers
Electrochemical reactions involve the transfer of electrons. Mass and charge are conserved when balancing these reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction. Oxidation numbers are used to keep track of how many electrons are lost or gained by each atom. These oxidation numbers are assigned using the following rules:
The convention is that the cation is written first in a formula, followed by the anion.For example, in NaH, the H is H-; in HCl, the H is H+.
The oxidation number of a free element is always 0.The atoms in He and N2, for example, have oxidation numbers of 0.
The oxidation number of a monatomic ion equals the charge of the ion.For example, the oxidation number of Na+ is +1; the oxidation number of N3- is -3.
The usual oxidation number of hydrogen is +1.The oxidation number of hydrogen is -1 in compounds containing elements that are less electronegative than hydrogen, as in CaH2.
The oxidation number of oxygen in compounds is usually -2.Exceptions include OF2, since F is more electronegative than O, and BaO2, due to the structure of the peroxide ion, which is [O-O]2-.
The oxidation number of a Group VIIA element in a compound is -1, except when that element is combined with one having a higher electronegativity.The oxidation number of Cl is -1 in HCl, but the oxidation number of Cl is +1 in HOCl.
The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.
The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.For example, the sum of the oxidation numbers for SO42- is -2.
Redox equations are often so complex that fiddling with coefficients to balance chemical equations doesn’t always work well. Chemists have developed an alternative method (in addition to the oxidation number method) that is called the ion-electron (half-reaction) method.
In the ion-electron method, the unbalanced redox equation is converted to the ionic equation and then broken down into two half-reactions — oxidation and reduction. Each of these half-reactions is balanced separately and then combined to give the balanced ionic equation.
Finally, the spectator ions are put into the balanced ionic equation, converting the reaction back to the molecular form. It’s important to follow the steps precisely and in the order listed. Otherwise, you may not be successful in balancing redox equations.
The example below shows you how to use the ion-electron method to balance this redox equation:
image0.png
Follow these steps:
Convert the unbalanced redox reaction to the ionic form.
In this reaction, you show the nitric acid in the ionic form, because it’s a strong acid. Copper(II) nitrate is soluble (indicated by (aq)), so it’s shown in its ionic form. Because NO(g) and water are molecular compounds, they remain shown in the molecular form:
image1.png
If necessary, assign oxidation numbers and then write two half-reactions (oxidation and reduction) showing the chemical species that have had their oxidation numbers changed.
In some cases, it’s easy to tell what has been oxidized and reduced; but in other cases, it isn’t as easy. Start by going through the example reaction and assigning oxidation numbers. You can then use the chemical species that have had their oxidation numbers changed to write your unbalanced half-reactions:
image2.png
Copper changed its oxidation number (from 0 to 2) and so has nitrogen (from –2 to +2). Your unbalanced half-reactions are:
image3.png
image4.png
Balance all atoms, with the exception of oxygen and hydrogen.
It’s a good idea to wait until the end to balance hydrogen and oxygen atoms, so always balance the other atoms first. You can balance them by fiddling with the coefficients. (You can’t change subscripts; you can only add coefficients.) However, in this particular case, both the copper and nitrogen atoms already balance, with one each on both sides:
image5.png
image6.png
Balance the oxygen atoms.
How you balance these atoms depends on whether you’re dealing with acid or basic solutions:
In acid solutions, take the number of oxygen atoms needed and add that same number of water molecules to the side that needs oxygen.
In basic solutions, add
image7.png
to the side that needs oxygen for every oxygen atom that is needed. Then, to the other side of the equation, add half as many water molecules as
image8.png
anions used.
The example equation is in acidic conditions. There’s nothing to do on the half-reaction involving the copper, because there are no oxygen atoms present. But you do need to balance the oxygen atoms in the second half-reaction:
image9.png
image10.png
Balance the hydrogen atoms.
Again, how you balance these atoms depends on whether you’re dealing with acid or basic solutions:
In acid solutions, take the number of hydrogen atoms needed and add that same number of
image11.png
to the side that needs hydrogen.
In basic solutions, add one water molecule to the side that needs hydrogen for every hydrogen atom that’s needed. Then, to the other side of the equation, add as many
image12.png
anions as water molecules used.
The example equation is in acidic conditions. You need to balance the hydrogen atoms in the second half-reaction:
image13.png
image14.png
Balance the ionic charge on each half-reaction by adding electrons.
The electrons should end up on opposite sides of the equation in the two half-reactions. Remember that you’re using ionic charge, not oxidation numbers.
Oxidation:
image15.png
Reduction:
image16.png
Balance electron loss with electron gain between the two half-reactions.
The electrons that are lost in the oxidation half-reaction are the same electrons that are gained in the reduction half-reaction. The number of electrons lost and gained must be the same. But Step 6 shows a loss of 2 electrons and a gain of 3.
So you must adjust the numbers using appropriate multipliers for both half-reactions. In this case, you have to find the lowest common denominator between 2 and 3. It’s 6, so multiply the first half-reaction by 3 and the second half-reaction by 2.
image17.png
image18.png
Add the two half-reactions together and cancel anything common to both sides.
The electrons should always cancel (the number of electrons should be the same on both sides).
image19.png
Convert the equation back to the molecular form by adding the spectator ions.
If it’s necessary to add spectator ions to one side of the equation, add the same number to the other side of the equation.
image20.png
Check to make sure that all the atoms are balanced, all the charges are balanced (if working with an ionic equation at the beginning), and all the coefficients are in the lowest whole-number ratio
Interpreting Electrochemical series
Consider the electrochemical series.
Consider the elements Zn > Cu> Ag The more active an element is the stronger the reductant.
So in this group Ag is the least reactive hence Zn is the strongest reductant in this group.
In the series above Au is the least reactive hence anything below it is a stronger reductant. and this means this elements below Au will be oxidised. OXIDATION will occur. [ remember AN OIL RIG CAT] .
So overall any element below Au will lose electrons (will be oxidised) before Au will lose electrons (be oxidised ) And we would find them at the anode giving off electrons while we would find reduction happening at the cathode gaining electrons.
Now this is also true for any element below another in this table.
So consider the series Zn > Cu > Ag
Cu is a stronger reductant than Ag ie Cu will be oxidised before Ag ie Cu will lose electrons before Ag. So if Cu and Ag were 2 half cells in a galvanic cell Cu will be losing the electrons ie oxidised ie at the anode and Ag will be reduced (ie the stronger oxidant) that is will gain electrons at the cathode .
Generally an element that is reduced will go from an ion to a solid - or work out the oxidation number and you should see the the oxidation number going down ie reduced.
In the electrochemical series
the strongest reductants are at the bottom of the table - this is the equivalent of saying oxidation occurs for elements below the others while reduction occurs for elements above the others.
The oxidant has been reduced and the reductant has been oxidised
An OIL RiG Cat
anode oxidation is lossReduction is gain cathode
Oxidation is loss of an electron (from an element) at the anode (this means the electrons are building up on this terminal )
eg
Cu(s) (reductant) -----> Cu 2+ (aq) + 2 e- electrons have been lost from the reactant, so now they appear as the product and the element has become a cation....the copper atom is oxidised
Reduction is gain of an electron at the cathode (this means the cation will gain an electron and become a neutral atom. For most neutral atoms of metals will be solid)
Ag+(aq) (oxidant) + e- ------> Ag(s)
(electron is gained so it appears as the reactant, it balances the charge on the ion)
(an electron is gained to reduce each silver ion)
The oxidant has been reduced and the reductant has been oxidised
Try this quizlet to practice the meanings
http://quizlet.com/26141502/flashcards
Cathode is the positive terminal
Anode is the negative terminal
These 2 equations can be added together
Cu(s) + 2Ag+(aq) -------> Cu2+(aq) + 2Ag(s)
Note that the silver equation has been multiplied by 2 so the electrons have been balanced out and hence cancelled out
Determining oxidation numbers
From ---http://www.chemguide.co.uk/inorganic/redox/oxidnstates.htmlThis page explains what oxidation states (oxidation numbers) are and how to calculate them and make use of them.
Oxidation states are straightforward to work out and to use, but it is quite difficult to define what they are in any quick way.
Explaining what oxidation states (oxidation numbers) are
Oxidation states simplify the whole process of working out what is being oxidised and what is being reduced in redox reactions. However, for the purposes of this introduction, it would be helpful if you knew about:
Note: If you aren't sure about either of these things, you might want to look at the pages on redox definitions and electron-half-equations. It would probably be best to read on and come back to these links if you feel you need to.
Vanadium forms a number of different ions - for example, V2+ and V3+. If you think about how these might be produced from vanadium metal, the 2+ ion will be formed by oxidising the metal by removing two electrons:
Removal of another electron gives the V3+ ion:
Removal of another electron gives a more unusual looking ion, VO2+.
The positive oxidation state is counting the total number of electrons which have had to be removed - starting from the element.
It is also possible to remove a fifth electron to give another ion (easily confused with the one before!). The oxidation state of the vanadium is now +5.
Fairly obviously, if you start adding electrons again the oxidation state will fall. You could eventually get back to the element vanadium which would have an oxidation state of zero.
What if you kept on adding electrons to the element? You can't actually do that with vanadium, but you can with an element like sulphur.
Summary
Oxidation state shows the total number of electrons which have been removed from an element (a positive oxidation state) or added to an element (a negative oxidation state) to get to its present state.
Oxidation involves an increase in oxidation state
Reduction involves a decrease in oxidation state
Recognising this simple pattern is the single most important thing about the concept of oxidation states. If you know how the oxidation state of an element changes during a reaction, you can instantly tell whether it is being oxidised or reduced without having to work in terms of electron-half-equations and electron transfers
.
Working out oxidation states
You don't work out oxidation states by counting the numbers of electrons transferred. It would take far too long. Instead you learn some simple rules, and do some very simple sums!
- The oxidation state of an uncombined element is zero. That's obviously so, because it hasn't been either oxidised or reduced yet! This applies whatever the structure of the element - whether it is, for example, Xe or Cl2 or S8, or whether it has a giant structure like carbon or silicon.
- The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
- The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
- The more electronegative element in a substance is given a negative oxidation state. The less electronegative one is given a positive oxidation state. Remember that fluorine is the most electronegative element with oxygen second.
- Some elements almost always have the same oxidation states in their compounds:
>The reasons for the exceptions
Hydrogen in the metal hydrides
Metal hydrides include compounds like sodium hydride, NaH. In this, the hydrogen is present as a hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion - in this case, -1.
Alternatively, you can think of it that the sum of the oxidation states in a neutral compound is zero. Since Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).
Oxygen in peroxides
Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound and so the sum of the oxidation states of the hydrogen and oxygen must be zero.
Since each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.
Oxygen in F2O
The problem here is that oxygen isn't the most electronegative element. The fluorine is more electronegative and has an oxidation state of -1. In this case, the oxygen has an oxidation state of +2.
Chlorine in compounds with fluorine or oxygen
There are so many different oxidation states that chlorine can have in these, that it is safer to simply remember that the chlorine doesn't have an oxidation state of -1 in them, and work out its actual oxidation state when you need it. You will find an example of this below.
Warning!
Don't get too bogged down in these exceptions. In most of the cases you will come across, they don't apply!Examples of working out oxidation states
What is the oxidation state of chromium in Cr2+?
That's easy! For a simple ion like this, the oxidation state is the charge on the ion - in other words: +2 (Don't forget the + sign.)
What is the oxidation state of chromium in CrCl3?
This is a neutral compound so the sum of the oxidation states is zero. Chlorine has an oxidation state of -1. If the oxidation state of chromium is n:
n + 3(-1) = 0
n = +3 (Again, don't forget the + sign!)
What is the oxidation state of chromium in Cr(H2O)63+?
This is an ion and so the sum of the oxidation states is equal to the charge on the ion. There is a short-cut for working out oxidation states in complex ions like this where the metal atom is surrounded by electrically neutral molecules like water or ammonia.
The sum of the oxidation states in the attached neutral molecule must be zero. That means that you can ignore them when you do the sum. This would be essentially the same as an unattached chromium ion, Cr3+. The oxidation state is +3.
What is the oxidation state of chromium in the dichromate ion, Cr2O72-?
The oxidation state of the oxygen is -2, and the sum of the oxidation states is equal to the charge on the ion. Don't forget that there are 2 chromium atoms present.
2n + 7(-2) = -2
n = +6
||
Warning: Because these are simple sums it is tempting to try to do them in your head. If it matters (like in an exam) write them down using as many steps as you need so that there is no chance of making careless mistakes. Your examiners aren't going to be impressed by your mental arithmetic - all they want is the right answer!
If you want some more examples to practice on, you will find them in most text books, including my chemistry calculations book.
Unfortunately, it isn't always possible to work out oxidation states by a simple use of the rules above. The problem in this case is that the compound contains two elements (the copper and the sulphur) whose oxidation states can both change.
The only way around this is to know some simple chemistry! There are two ways you might approach it. (There might be others as well, but I can't think of them at the moment!)
- You might recognise this as an ionic compound containing copper ions and sulphate ions, SO42-. To make an electrically neutral compound, the copper must be present as a 2+ ion. The oxidation state is therefore +2.
- You might recognise the formula as being copper(II) sulphate. The "(II)" in the name tells you that the oxidation state is 2 (see below).
Using oxidation statesYou will know that it is +2 because you know that metals form positive ions, and the oxidation state will simply be the charge on the ion.
In naming compounds
You will have come across names like iron(II) sulphate and iron(III) chloride. The (II) and (III) are the oxidation states of the iron in the two compounds: +2 and +3 respectively. That tells you that they contain Fe2+ and Fe3+ ions.
This can also be extended to the negative ion. Iron(II) sulphate is FeSO4. There is also a compound FeSO3 with the old name of iron(II) sulphite. The modern names reflect the oxidation states of the sulphur in the two compounds.
The sulphate ion is SO42-. The oxidation state of the sulphur is +6 (work it out!). The ion is more properly called the sulphate(VI) ion.
The sulphite ion is SO32-. The oxidation state of the sulphur is +4 (work that out as well!). This ion is more properly called the sulphate(IV) ion. The ate ending simply shows that the sulphur is in a negative ion.
So FeSO4 is properly called iron(II) sulphate(VI), and FeSO3 is iron(II) sulphate(IV). In fact, because of the easy confusion between these names, the old names sulphate and sulphite are normally still used in introductory chemistry courses.
Note: Even these aren't the full name! The oxygens in the negative ions should also be identified. FeSO4 is properly called iron(II) tetraoxosulphate(VI). It all gets a bit out of hand for everyday use for common ions.
This is easily the most common use of oxidation states.
Remember:
Oxidation involves an increase in oxidation state
Reduction involves a decrease in oxidation state
In each of the following examples, we have to decide whether the reaction involves redox, and if so what has been oxidised and what reduced.
Example 1:
This is the reaction between magnesium and hydrochloric acid or hydrogen chloride gas:
Example 2:
The reaction between sodium hydroxide and hydrochloric acid is:
Example 3:
This is a sneaky one! The reaction between chlorine and cold dilute sodium hydroxide solution is:
This is a good example of a disproportionation reaction. A disproportionation reaction is one in which a single substance is both oxidised and reduced.
Using oxidation states to work out reacting proportions
This is sometimes useful where you have to work out reacting proportions for use in titration reactions where you don't have enough information to work out the complete ionic equation.
Remember that each time an oxidation state changes by one unit, one electron has been transferred. If one substance's oxidation state in a reaction falls by 2, that means that it has gained 2 electrons.
Something else in the reaction must be losing those electrons. Any oxidation state fall by one substance must be accompanied by an equal oxidation state increase by something else.
This example is based on information in an old AQA A' level question.
Ions containing cerium in the +4 oxidation state are oxidising agents. (They are more complicated than just Ce4+.) They can oxidise ions containing molybdenum from the +2 to the +6 oxidation state (from Mo2+ to MoO42-). In the process the cerium is reduced to the +3 oxidation state (Ce3+). What are the reacting proportions?
The oxidation state of the molybdenum is increasing by 4. That means that the oxidation state of the cerium must fall by 4 to compensate.
But the oxidation state of the cerium in each of its ions only falls from +4 to +3 - a fall of 1. So there must obviously be 4 cerium ions involved for each molybdenum ion.
The reacting proportions are 4 cerium-containing ions to 1 molybdenum ion.
Or to take a more common example involving iron(II) ions and manganate(VII) ions . . .
A solution of potassium manganate(VII), KMnO4, acidified with dilute sulphuric acid oxidises iron(II) ions to iron(III) ions. In the process, the manganate(VII) ions are reduced to manganese(II) ions. Use oxidation states to work out the equation for the reaction.
The oxidation state of the manganese in the manganate(VII) ion is +7. The name tells you that, but work it out again just for the practice!
In going to manganese(II) ions, the oxidation state of manganese has fallen by 5. Every iron(II) ion that reacts, increases its oxidation state by 1. That means that there must be five iron(II) ions reacting for every one manganate(VII) ion.
The left-hand side of the equation will therefore be: MnO4- + 5Fe2+ + ?
The right-hand side will be: Mn2+ + 5Fe3+ + ?
After that you will have to make guesses as to how to balance the remaining atoms and the charges. In this case, for example, it is quite likely that the oxygen will end up in water. That means that you need some hydrogen from somewhere.
That isn't a problem because you have the reaction in acid solution, so the hydrogens could well come from hydrogen ions.
Eventually, you will end up with this:
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© Jim Clark 2002 (last modified February 2013)
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(from http://chemistry.about.com/od/generalchemistry/a/oxidationno.htm)
Electrochemical reactions involve the transfer of electrons. Mass and charge are conserved when balancing these reactions, but you need to know which atoms are oxidized and which atoms are reduced during the reaction. Oxidation numbers are used to keep track of how many electrons are lost or gained by each atom. These oxidation numbers are assigned using the following rules:
Suggested Reading
How to Balance Redox Equations
Redox equations are often so complex that fiddling with coefficients to balance chemical equations doesn’t always work well. Chemists have developed an alternative method (in addition to the oxidation number method) that is called the ion-electron (half-reaction) method.In the ion-electron method, the unbalanced redox equation is converted to the ionic equation and then broken down into two half-reactions — oxidation and reduction. Each of these half-reactions is balanced separately and then combined to give the balanced ionic equation.
Finally, the spectator ions are put into the balanced ionic equation, converting the reaction back to the molecular form. It’s important to follow the steps precisely and in the order listed. Otherwise, you may not be successful in balancing redox equations.
The example below shows you how to use the ion-electron method to balance this redox equation:
Follow these steps:
In this reaction, you show the nitric acid in the ionic form, because it’s a strong acid. Copper(II) nitrate is soluble (indicated by (aq)), so it’s shown in its ionic form. Because NO(g) and water are molecular compounds, they remain shown in the molecular form:
In some cases, it’s easy to tell what has been oxidized and reduced; but in other cases, it isn’t as easy. Start by going through the example reaction and assigning oxidation numbers. You can then use the chemical species that have had their oxidation numbers changed to write your unbalanced half-reactions:
It’s a good idea to wait until the end to balance hydrogen and oxygen atoms, so always balance the other atoms first. You can balance them by fiddling with the coefficients. (You can’t change subscripts; you can only add coefficients.) However, in this particular case, both the copper and nitrogen atoms already balance, with one each on both sides:
How you balance these atoms depends on whether you’re dealing with acid or basic solutions:
- In acid solutions, take the number of oxygen atoms needed and add that same number of water molecules to the side that needs oxygen.
- In basic solutions, add
image7.png
to the side that needs oxygen for every oxygen atom that is needed. Then, to the other side of the equation, add half as many water molecules as
image8.png
anions used.
The example equation is in acidic conditions. There’s nothing to do on the half-reaction involving the copper, because there are no oxygen atoms present. But you do need to balance the oxygen atoms in the second half-reaction:Again, how you balance these atoms depends on whether you’re dealing with acid or basic solutions:
- In acid solutions, take the number of hydrogen atoms needed and add that same number of
image11.png
to the side that needs hydrogen. - In basic solutions, add one water molecule to the side that needs hydrogen for every hydrogen atom that’s needed. Then, to the other side of the equation, add as many
image12.png
anions as water molecules used.
The example equation is in acidic conditions. You need to balance the hydrogen atoms in the second half-reaction:The electrons should end up on opposite sides of the equation in the two half-reactions. Remember that you’re using ionic charge, not oxidation numbers.
Oxidation:
The electrons that are lost in the oxidation half-reaction are the same electrons that are gained in the reduction half-reaction. The number of electrons lost and gained must be the same. But Step 6 shows a loss of 2 electrons and a gain of 3.
So you must adjust the numbers using appropriate multipliers for both half-reactions. In this case, you have to find the lowest common denominator between 2 and 3. It’s 6, so multiply the first half-reaction by 3 and the second half-reaction by 2.
The electrons should always cancel (the number of electrons should be the same on both sides).
If it’s necessary to add spectator ions to one side of the equation, add the same number to the other side of the equation.
Interpreting Electrochemical series
Consider the electrochemical series.
Consider the elements Zn > Cu> Ag The more active an element is the stronger the reductant.
So in this group Ag is the least reactive hence Zn is the strongest reductant in this group.
In the series above Au is the least reactive hence anything below it is a stronger reductant. and this means this elements below Au will be oxidised. OXIDATION will occur. [ remember AN OIL RIG CAT] .
So overall any element below Au will lose electrons (will be oxidised) before Au will lose electrons (be oxidised ) And we would find them at the anode giving off electrons while we would find reduction happening at the cathode gaining electrons.
Now this is also true for any element below another in this table.
So consider the series Zn > Cu > Ag
Cu is a stronger reductant than Ag ie Cu will be oxidised before Ag ie Cu will lose electrons before Ag. So if Cu and Ag were 2 half cells in a galvanic cell Cu will be losing the electrons ie oxidised ie at the anode and Ag will be reduced (ie the stronger oxidant) that is will gain electrons at the cathode .
Generally an element that is reduced will go from an ion to a solid - or work out the oxidation number and you should see the the oxidation number going down ie reduced.
In the electrochemical series
the strongest reductants are at the bottom of the table - this is the equivalent of saying oxidation occurs for elements below the others while reduction occurs for elements above the others.
The oxidant has been reduced and the reductant has been oxidised